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GUJCET · Maths · Inverse Trigonometric Functions
\(\cot ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\) ____________
- A \(-\frac{1}{2} \tan ^{-1} x\)
- B \(\cot ^{-1} x\)
- C \(\frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x\)
- D \(\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x\)
Step-by-step Solution
Detailed explanation
Let \(x = \tan \theta\). \(\cot ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) = \cot ^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right) = \cot ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)\)
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